Question : If the sum of the diagonals of a rhombus is $L$ and the perimeter is $4P$, find the area of the rhombus.
Option 1: $\frac{1}{4}\left(\mathrm{~L}^2-\mathrm{P}^2\right)$
Option 2: $\frac{1}{2}\left(\mathrm{~L}^2-4 \mathrm{P}^2\right)$
Option 3: $\frac{1}{4}\left(\mathrm{~L}^2+3 \mathrm{P}^2\right)$
Option 4: $\frac{1}{4}\left(\mathrm{~L}^2-4 \mathrm{P}^2\right)$
Correct Answer: $\frac{1}{4}\left(\mathrm{~L}^2-4 \mathrm{P}^2\right)$
Solution :
Perimeter = $4P$
⇒ 4 × side = $4P$
⇒ Side = $P$
Let $d_1$ and $d_2$ be the diagonals.
The sum of the diagonals = $L$
⇒ $d_1+d_2$ = L ----------------------(i)
Now, (Side)$^2$ = $(\frac{d_1}{2})^2 + (\frac{d_2}{2})^2$
⇒ $P^2 = (\frac{d_1}{2})^2 + (\frac{d_2}{2})^2$
⇒ $d_1^2+d_2^2 = 4P^2$
⇒ $(d_1+d_2)^2-2d_1d_2 = 4P^2$
⇒ $L^2-2d_1d_2 = 4P^2$
⇒ $d_1d_2 = \frac{L^2-4P^2}{2}$
So, the area of the rhombus = $\frac{1}{2}d_1d_2 = \frac{L^2-4P^2}{4}$
Hence, the correct answer is $\frac{1}{4}(L^2-4P^2)$.
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